I am trying to understand *Remark 7.2.22 (Page 256)* of *Algebraic Function Fields and Codes* (Second Edition) by *Henning Stichtenoth*.

In that remark he considers a tower $\mathcal{F} = (F_0,F_1,F_2,\dots)$ over a finite constant field $\mathbb{F_q}$ and extends it constantly with $L$ to get $\mathcal{F'}=(F_0',F_1',F_2',\dots)$ (i.e. $L/\mathbb{F_q}$ is an algebraic extension (may be finite or infinite) and $F_i' = F_iL$ $\forall i \geq 1$).

Next he takes a place

$P \in \mathbb{P}(F_i)$ for some $i >0$ and asserts that $P$ ramifies in the extension $F_{i+1}/F_i$ if and only if the places $P' \in \mathbb{P}(F_i')$ above $P$ are ramified in the extension $F_{i+1}'/F_i'$.

My doubt here is if the situation is as given below

**Case 1:**
$$
\begin{array}{ccccccc}
F_i' & \rightarrow & F_{i+1}' &&P'&\rightarrow & P_1'\\
\uparrow & & \uparrow && \uparrow && \uparrow\\
F_i& \rightarrow & F_{i+1} && P &\xrightarrow{e>1} & P_1
\end{array}
$$
(where $e$ is the corresponding ramification index in the diagram) then as $e(P'/P)=e(P_1'/P_1)=1$ (because $F_i'/F_i$ is a constant field extension) we can say that $e(P_1'/P')=e(P_1/P) >1$.

But what if the situation as depicted in the following diagram also happens?

**Case 2:**
$$
\begin{array}{ccc}
Q & \rightarrow & P_2'\\
\uparrow && \uparrow \\
P & \xrightarrow{e=1} & P_2
\end{array}
$$
i.e. $P_1, P_2$ both extend $P$ with one extension ramified and another unramified. In this case $e(P_2'|Q)$ has to be equal to $1$.

How can he conclude that every place $P' \in \mathbb{P}(F_i')$ gets ramified if $P$ is ramified? Doesn't **Case 2** situation arise?

p.s. I have asked the same question in math.stackexchange.com but I could not get an answer there (update: now there's an answer).